Question

Find the inverse point of P(-2,3) with the circle x²+y²-4x-6y+9=0​

Answer 1

The inverse point of P(-2,3) with the circle x²+y²-4x-6y+9=0​ is (1,3)

Step-by-step explanation:

The given equation of the circle is

$x^2+y^2-4x-6y+9=0$

$\implies (x^2-2\times2x+4)+(y^2-2\times 3y+9)+9-4-9=0$

$\implies x^2-2\times x\times 2+2^2)+(y^2-2\times y]times 3+3^2)=4$

$\implies (x-2)^2+(y-3)^2=2^2$

Clearly, we can say that the centre of the circle is $(2,3)$ and radius is $2$ units

We know that inverse of a point P (x',y') w.r.t. a circle with centre at (h,k) and radius r is point (x",y") such that

$x"=\alpha(x'-h)+h$

And, $y"=\alpha(y'-k)+k$

Where,

$\alpha=\frac{r^2}{(x'-h)^2+(y'-k)^2}$

The given point is (-2,3)

Thus,

$\alpha=\frac{4}{(-2-2)^2+(3-3)^3}$

$\implies \alpha=\frac{1}{4}$

Therefore,

the x-coordinate of the inverse point

$x"=\frac{1}{4}(-2-2)+2$

$\implies x"=1$

The y-coordinate of the inverse point

$x"=\frac{1}{4}(3-3)+3$

$\implies y"=3$

Therefore, the coordinate of the inverse point is $(1,3)$

Hope this answer is helpful.

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Q: The inverse of (1,3) with the respect of the circle x2+y2-4x-6y+9=0

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